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RSB [31]
3 years ago
10

What is real-life relationship that might be described by the graph?

Mathematics
2 answers:
oee [108]3 years ago
4 0

Answer:

The real-life relationship that might be described by the graph is the number of days in a week.

Step-by-step explanation:

The graph represents a linear function. Linear functions are functions that have a constant rate of change and are expressed as.

y=mx +b

where x and y are the independent and dependent variables , m is the constant rate of change and b is the intercept of the function in the y chord. In your case taking time weeks as x  (W) variable and time days as y variable (D)

D = mW + b

when week is 0 eh graph shows that D=0 so

0 =  m(0) + b

0 = b

b= 0

m represent how much the y variable changes when x increases by one. Any week has 7 days and this relationship is constant no matter how many weeks we are talking about so

1 week  - 7 days

so m = 7. One week corresponds to 7 days

and the function can be represented as

D= 7W

Luden [163]3 years ago
3 0
This specific graph really has no real life application and is just used to demonstrate how graphs work. However graphs can be useful when trying to model a system in real life. For example you can use a graph to model the relationship between altitude and temperature. You could also model the relationship between hours studied and grade received on a test. Graphs are useful because they allow you to visually interpret data, and make more highly educated predictions about the behavior of a system.
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3 years ago
How do I solve this equation?
svlad2 [7]
The answer is option three(photo for step by step)

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3 years ago
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If m/CHF = 106°, then what is m/FHD?
densk [106]

Answer:

74 degrees

Step-by-step explanation:

The two angles in question form a straight line (180 degrees)

   180 = CHF + FHD

  180 = 106 + FHD

180 - 106 = FHD

   FHD = 74 degrees

5 0
2 years ago
The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional baseball players gav
Andru [333]

Step-by-step explanation:

(a) Yes, if you enter all 43 values into your calculator, you calculator should report:

xbar = 2.293

s = 1.401

(b)

Note: Most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 1.684 * 1.401 / sqrt(43) = 1.933

2.293 - 1.684 * 1.401 / sqrt(43) = 2.653

Answer: (1.933, 2.653)

Note: To find the t-value that allows us to be 90% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.90)/2 = .05 or up from 90% depending on your t-table. So, the t-critical value is 1.684.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 1.681).

2.293 +/- 1.681 * 1.401 / sqrt(43)

(1.934, 2.652)

Note: Some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 90% CI is:

2.293 +/- 1.645 * 1.401 / sqrt(43)

(1.942, 2.644)

Note: To find the z-value that allows us to be 90% confident, (1) using the z-table, look up (1-.90)/2 = .05 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .05 or up from 90% depending on your t-table. Either way, the z-critical value is 1.645.

(c)

Note: Again, most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 2.704 * 1.401 / sqrt(43) = 1.715

2.293 - 2.704 * 1.401 / sqrt(43) = 2.871

Answer: (1.715, 2.871)

Note: To find the t-value that allows us to be 99% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.99)/2 = .005 or up from 99% depending on your t-table. So, the t-critical value is 2.704.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 2.698).

2.293 +/- 2.698 * 1.401 / sqrt(43)

(1.717, 2.869)

Note: Again, some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 99% CI is:

2.293 +/- 2.576 * 1.401 / sqrt(43)

(1.742, 2.843)

Note: To find the z-value that allows us to be 99% confident, (1) using the z-table, look up (1-.99)/2 = .005 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .005 or up from 99% depending on your t-table. Either way, the z-critical value is 2.576.

(d)

Tim Huelett 2.5

Since 2.5 falls between (1.715, 2.871), we see that Tim Huelett falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Herb Hunter 2.0

Since 2.0 falls between (1.715, 2.871), we see that Herb Hunter falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Jackie Jensen 3.8.

Since 3.8 falls above (1.715, 2.871), we see that Jackie Jensen falls in the 99% CI range. So, his home run percentage IS significantly GREATER than the population average.

(e)

Because of the Central Limit Theorem (CLT), since our sample size is large, we do NOT have to make the normality assumption since the CLT tells us that the sampling distribution of xbar will be approximatley normal even if the underlying population distribution is not.

6 0
3 years ago
2(3x + 4) + 6 + 7x<br> What is the answer
Colt1911 [192]

Answer:

13x+14

Step-by-step explanation:

2(3x+4)+6+7x

Distribute:

=(2)(3x)+(2)(4)+6+7x

=6x+8+6+7x

Combine Like Terms:

=6x+8+6+7x

=(6x+7x)+(8+6)

=13x+14

3 0
3 years ago
Read 2 more answers
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